Let's consider the following:
- for $d \geq 1$, $\mathbb{P}$ is the Lebesgue measure on $[0,1]^d$
- for $n \geq 0,k_1,\ldots,k_d \in \{0,\ldots,2^n-1 \}$ let $$ Q_n(k_1,\ldots,k_d) = \prod_{j = 1}^d \left( \frac{k_j}{2^n}, \frac{k_j + 1}{2^n} \right] $$
- for $f \in L^1(\mathbb{P})$ and $x \in [0,1]^d$
Let $(f_n)_{n \in \mathbb{N}}$ be the process defined as: $$ f_n(x) := 2^{nd} \int_{Q_n(k_1,\ldots,k_n)} f(y) \text{d}\mathbb{P}(y) $$ whenever $x \in Q_n(k_1,\ldots,k_d)$.
I've already proved that $(f_n)_{n \in \mathbb{N}}$ is a martingale with respect to the filtration $$ \mathcal{F}_n = \sigma \left( Q_n(k_1,\ldots,k_d) \colon k_1,\ldots,k_d \in \{0,\ldots,2^n-1\} \right). $$
Now I have to prove that $\lim_{n \to \infty} f_n = f$ in $L^1$.
I tried to show that $(f_n)$ is uniform integrable unsuccessfully. Even using the definition of $L^1$ convergence I can't go anywhere.
Any hint would be appreciated.
Notice that $f_n=\mathbb E\left[f\mid\mathcal F_n\right]$ hence uniform integrability of $(f_n)$ follows from the fact that if $X$ is an integrable random variable, then the collection $\left\{\mathbb E[X\mid\mathcal G],\mathcal G\mbox{ is a sub-}\sigma-\mbox{algebra of }\mathcal F\right\}$ is uniformly integrable. This has been discussed over this website. One can use de la Vallée Poussin theorem or truncate $X$.
The martingale convergence theorem guarantees that $f_n\to f_\infty$ in $\mathbb L^1$ for some random variable which is measurable with respect to $\mathcal F_\infty:=\sigma\left(\bigcup_{n=1}^\infty\mathcal F_n\right)$. We also have $f_n=\mathbb E\left[f_\infty\mid\mathcal F_n\right]$ hence for each $n$ and each $A\in\mathcal F_n$, $$\tag{*} \mathbb E\left[\left(f_\infty -f\right)\mathbb{1}_A\right]=0. $$ Approximating a set $A$ in $\mathcal F_\infty$ by a set in $\mathcal F_n$ for some $n$, one get (*) for $A\in\mathcal F_\infty$ hence $f_\infty=\mathbb E[f\mid\mathcal F_\infty]$. Then it remains to notice that $\mathcal F_\infty$ is the Borel $\sigma$-algebra.